Lecture 1 & 2
Kinematics equation, cont.
When an object moves with constant acceleration, the instantaneous acceleration
at any point in a time interval is equal to the value of the average acceleration
over the entire time interval.
So, we can write a =
then
Kinematics equation, cont.
Kinematics equation, cont.
Problem:
A golf ball is released from rest at the top of a very tall building. Neglecting air
resistance, calculate the position and velocity of the ball after 1.00 s, 2.00 s,
and 3.00 s.
Solution:
t (s)
v (m/s)
y (m)
1.00
2.00
3.00
-9.8
-19.6
-29.4
- 4.9
-19.6
- 44.1
Remarks: The minus signs on v mean that the velocity vectors are directed
downward, while the minus signs on y indicates positions below the origin.
The velocity of a falling object is directly proportional to the time, and the
position is proportional to the time squared, results first proven by Galileo.
Home work:
1. Calculate the position and velocity of the ball after 4.00 s
has elapsed.
2. If the velocity of a particle is nonzero, can the particle’s
acceleration be zero? Explain.
3. If the velocity of a particle is zero, can the particle's
acceleration be zero? Explain.
4. 4. If a car is traveling eastward, can its acceleration be
westward? Explain.